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\(\left(\frac{1}{2}\right)^5\times x=\left(\frac{1}{2}\right)^7\)
\(x=\left(\frac{1}{2}\right)^7\div\left(\frac{1}{2}\right)^5\)
\(x=\left(\frac{1}{2}\right)^{7-5}=\left(\frac{1}{2}\right)^2=\frac{1}{4}\) .
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{9}{21}\right)^2\)
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{3}{7}\right)^4\)
\(x=\left(\frac{3}{7}\right)^4\div\left(\frac{3}{7}\right)^2\)
\(x=\left(\frac{3}{7}\right)^{4-2}=\left(\frac{3}{7}\right)^2=\frac{9}{49}\)
\(2^x=2\Rightarrow x=1\)
\(3^x=3^4\Rightarrow x=4\)
\(7^x=7^7\Rightarrow x=7\)
\(\left(-3\right)^x=\left(-3\right)^5\Rightarrow x=5\)
\(\left(-5\right)^x=\left(-5\right)^4\Rightarrow x=4\)
\(2^x=4\Leftrightarrow2^x=2^2\Rightarrow x=2\)
\(2^x=8\Leftrightarrow2^x=2^3\Rightarrow x=3\)
\(2^x=16\Leftrightarrow2^x=2^4\Rightarrow x=4\)
\(3^{x+1}=3^2\Leftrightarrow x+1=2\Leftrightarrow x=2-1\Rightarrow x=1\)
\(5^{x-1}=5\Leftrightarrow x-1=1\Leftrightarrow x=1+1\Rightarrow x=2\)
\(6^{x+4}=6^{10}\Leftrightarrow x+4=10\Leftrightarrow x=10-4\Rightarrow x=6\)
\(5^{2x-7}=5^{11}\Leftrightarrow2x-7=11\Leftrightarrow2x=11+7\Leftrightarrow2x=18\Leftrightarrow x=18\div2\Rightarrow x=9\)
\(\left(-2\right)^{4x+2}=64\)
\(2^{-4x+2}=2^6\Leftrightarrow-4x+2=6\Leftrightarrow-4x=6-2\Leftrightarrow-4x=4\Leftrightarrow x=4\div\left(-4\right)\Rightarrow x=-1\)
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^5\Rightarrow x=5\)
\(\left(\frac{5}{6}\right)^{2x}=\left(\frac{5}{6}\right)^5\Rightarrow2x=5\Rightarrow x=\frac{5}{2}\)
\(\left(\frac{3}{4}\right)^{2x-1}=\left(\frac{3}{4}\right)^{5x-4}\Rightarrow2x-1=5x-4\)
\(2x-5x=-4+1\)
\(-3x=-3\Rightarrow x=1\)
\(\left(\frac{-1}{10}\right)^x=\frac{1}{100}\)
\(\left(\frac{1}{10}\right)^{-x}=\left(\frac{1}{10}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{2}\right)^x=\frac{9}{4}\)
\(\left(\frac{3}{2}\right)^{-x}=\left(\frac{3}{2}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{5}\right)^{2x}=\frac{9}{25}\)
\(\left(\frac{3}{5}\right)^{-2x}=\left(\frac{3}{5}\right)^2\Rightarrow-2x=2\Rightarrow x=-1\)
\(\left(\frac{-2}{3}\right)^x=\frac{-8}{27}\)
\(\left(\frac{-2}{3}\right)^x=\left(\frac{-2}{3}\right)^3\Rightarrow x=3\).
hehe.
đánh tới què tay, hoa mắt lun r nekkk!!
\(\left(\frac{5}{x+3}-2\right).4=7-\left(\frac{9}{x+3}+\frac{1}{2}\right).2\)
\(\Leftrightarrow\frac{20}{x+3}-8=7-\frac{18}{x+3}+1\)
\(\Leftrightarrow\frac{20}{x+3}-8=8-\frac{18}{x+3}\)
\(\Leftrightarrow\frac{20}{x+3}+\frac{18}{x+3}=8+8\)
\(\Leftrightarrow\frac{38}{x+3}=16\)
\(\Leftrightarrow x+3=2,375\)
\(\Leftrightarrow x=-0,625\)
\(\left(\frac{5}{x+3}-2\right).4=7-\left(\frac{9}{x+3}+\frac{1}{2}\right).2\)
\(\Leftrightarrow\frac{20}{x+3}-8=7-\left(\frac{18}{x+3}+1\right)\)
\(\Leftrightarrow\frac{20}{x+3}-8=7-\frac{18}{x+3}-1\)
\(\Leftrightarrow\frac{20}{x+3}+\frac{18}{x+3}=7-1+8\)
\(\Leftrightarrow\frac{38}{x+3}=14\)
\(\Leftrightarrow\left(x+3\right)14=38\)
\(\Leftrightarrow14x+42=38\)
\(\Leftrightarrow14x=-4\Leftrightarrow x=-\frac{4}{14}=-\frac{2}{7}\)
Vậy \(x=-\frac{2}{7}\)
a) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
=> \(\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)
=> \(\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}+\frac{x+1}{12}=0\)
=> \(\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)
=> x + 1 = 0
=> x = -1
b) \(\frac{x-1}{2020}+\frac{x-2}{2019}-\frac{x-3}{2018}=\frac{x-4}{2017}\)
=> \(\left(\frac{x-1}{2020}-1\right)+\left(\frac{x-2}{2019}-1\right)-\left(\frac{x-3}{2018}-1\right)=\left(\frac{x-4}{2017}-1\right)\)
=> \(\frac{x-2021}{2020}+\frac{x-2021}{2019}-\frac{x-2021}{2018}=\frac{x-2021}{2017}\)
=> \(\left(x-2021\right)\left(\frac{1}{2020}+\frac{1}{2019}-\frac{1}{2018}-\frac{1}{2017}\right)=0\)
=> x - 2021 = 0
=> x = 2021
c) \(\left(\frac{3}{4}x+3\right)-\left(\frac{2}{3}x-4\right)-\left(\frac{1}{6}x+1\right)=\left(\frac{1}{3}x+4\right)-\left(\frac{1}{3}x-3\right)\)
=> \(\frac{3}{4}x+3-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}x+3\)
=> \(-\frac{1}{12}x+6=7\)
=> \(-\frac{1}{12}x=1\)
=> x = -12
\(\Leftrightarrow2.\left(\frac{-1}{2}\right).\left(\frac{2}{3}\right)^2-3\left(-\frac{1}{3}\right)^2.\frac{2}{9}:x=3.\left(-\frac{1}{2}\right)-\frac{2}{3}\)
\(\Leftrightarrow-\frac{4}{9}-\frac{1}{3}.\frac{2}{9}:x=-\frac{3}{2}-\frac{2}{3}\)
\(\Leftrightarrow-\frac{4}{6}-\frac{2}{27}:x=-\frac{13}{6}\)
\(\Leftrightarrow\frac{2}{27}:x=-\frac{4}{9}:\frac{-13}{6}\)
\(\Leftrightarrow\frac{2}{27}:x=\frac{31}{18}\)
\(\Leftrightarrow x=\frac{2}{27}:\frac{31}{18}\)
\(\Rightarrow x=\frac{4}{93}\)
Vậy \(x=\frac{4}{93}\)
a ) \(3-4.\left|5-6x\right|=7\)
\(\Leftrightarrow4.\left|5-6x\right|=-4\)
\(\Leftrightarrow\left|5-6x\right|=-1\)
\(\Leftrightarrow\) Không thõa mãn ( vì \(x\ge0\) )
b) Do \(\left|x+2\right|\ge0;\left|x+\frac{3}{5}\right|\ge0;\left|x+\frac{1}{2}\right|\ge0\)
=> \(4x\ge0\)
=> \(x\ge0\)
Lúc này ta có: \(\left(x+2\right)+\left(x+\frac{3}{5}\right)+\left(x+\frac{1}{2}\right)=4x\)
=> \(\left(x+x+x\right)+\left(2+\frac{3}{5}+\frac{1}{2}\right)=4x\)
=> \(3x+\frac{31}{10}=4x\)
=> \(4x-3x=\frac{31}{10}\)
=> \(x=\frac{31}{10}\)
Vậy \(x=\frac{31}{10}\)
c) Do \(\left|x+\frac{1}{101}\right|\ge0;\left|x+\frac{2}{101}\right|\ge0;\left|x+\frac{3}{101}\right|\ge0;...;\left|x+\frac{100}{101}\right|\ge0\)
=> \(101x\ge0\)
=> \(x\ge0\)
Lúc này ta có: \(\left(x+\frac{1}{101}\right)+\left(x+\frac{2}{101}\right)+\left(x+\frac{3}{101}\right)+...+\left(x+\frac{100}{101}\right)=101x\)
=> \(\left(x+x+x+...+x\right)+\left(\frac{1}{101}+\frac{2}{101}+\frac{3}{101}+...+\frac{100}{101}\right)=101x\)
100 số x
=> \(100x+\frac{\left(1+100\right).100:2}{101}=101x\)
=> \(\frac{101.50}{101}=101x-100x\)
=> \(x=50\)
Vậy x = 50
a, \(\left(\frac{1}{2}-\frac{1}{3}\right)\cdot6^x+6^{x+2}=6^{10}+6^7\)
\(\Leftrightarrow\frac{1}{6}\cdot6^x+6^x\cdot6^2=6^{10}+6^7\)
\(\Leftrightarrow6^{x-1}\left(1+6^3\right)=6^7\left(6^3+1\right)\)
\(\Leftrightarrow6^{x-1}=6^7\Leftrightarrow x-1=7\)
\(\Leftrightarrow x=8\)
b, \(\left(\frac{1}{2}-\frac{1}{6}\right)\cdot3^{x+4}-4\cdot3^x=3^{16}-4\cdot3^{13}\)
\(\Leftrightarrow\frac{1}{3}\cdot3^{x+4}-4\cdot3^x=3^{13}\left(3^3-4\right)\)
\(\Leftrightarrow3^x\cdot3^3-4\cdot3^x=3^{13}\left(3^3-4\right)\)
\(\Leftrightarrow3^x\left(3^3-4\right)=3^{13}\left(3^3-4\right)\)
\(\Leftrightarrow3^x=3^{13}\Leftrightarrow x=13\)
a. x=8
b. x=13
còn cách tính thì mình quên rồi vì minh học cái này lâu lắm rồi ko nhớ đc.
\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(\Rightarrow-\frac{13}{3}.\left(\frac{3}{6}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{4}{12}-\frac{6}{12}-\frac{9}{12}\right)\)
\(\Rightarrow-\frac{13}{3}.\frac{2}{6}\le x\le-\frac{2}{3}.\frac{-11}{12}\)
\(\Rightarrow\frac{-13}{9}\le x\le\frac{11}{18}\)
\(\Rightarrow\frac{-26}{18}\le x\le\frac{11}{18}\)
=> -1,44444444444........... ≤ x ≤ 0,6111111111...........
Mà x ∈ Z
=> x ∈ { -1 ; 0 }
\(a,5,5-\left|x-0,4\right|=-1\frac{1}{5}\)
\(\Rightarrow5,5-\left|x-0,4\right|=-\frac{6}{5}\)
\(\Rightarrow-\left|x-0,4\right|=-\frac{6}{5}-5,5=-6,7\)
\(\Rightarrow\left|x-0,4\right|=6,7\)
\(\Rightarrow x-0,4=\pm6,7\)
\(\Rightarrow\orbr{\begin{cases}x-0,4=6,7\\x-0,4=-6,7\end{cases}\Rightarrow\orbr{\begin{cases}x=7,1\\x=-6,3\end{cases}}}\)
\(a,5,5-\left|x-0,4\right|=-1\frac{1}{5}\)
=> \(\left|x-0,4\right|=5,5-\left[-\frac{6}{5}\right]=5,5+1,2=6,7\)
=> \(\left|x-0,4\right|=\pm6,7\)
Xét hai trường hợp :
TH1 : x - 0,4 = 6,7
=> x = 6,7 + 0,4 = 7,1
TH2 : x - 0,4 = -6,7
=> x = -6,7 + 0,4 =-6,3
\(b,\left[1-\frac{3}{4}\left|x\right|\right]^2=\frac{16}{25}\)
=> \(\left[1-\frac{3}{4}\left|x\right|\right]=\pm\sqrt{\frac{16}{25}}\)
=> \(\left[1-\frac{3}{4}\left|x\right|\right]=\pm\frac{4}{5}\)
=> \(\orbr{\begin{cases}1-\frac{3}{4}\left|x\right|=\frac{4}{5}\\1-\frac{3}{4}\left|x\right|=-\frac{4}{5}\end{cases}}\)=> \(\orbr{\begin{cases}x=\pm\frac{4}{15}\\x=\pm\frac{12}{5}\end{cases}}\)
\(c,\left[0,1\left|x\right|-\frac{1}{2}\right]\left[0,5-\left|x\right|\right]=0\)
=> \(\orbr{\begin{cases}0,1\left|x\right|-\frac{1}{2}=0\\0,5-\left|x\right|=0\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{1}{10}\left|x\right|=\frac{1}{2}\\\left|x\right|=0,5\end{cases}}\)
=> \(\orbr{\begin{cases}\left|x\right|=5\\\left|x\right|=0,5\end{cases}}\)=> \(\orbr{\begin{cases}x\in\left\{5;-5\right\}\\x\in\left\{0,5;-0,5\right\}\end{cases}}\)
d, Xét hai trường hợp rồi ra kết quả thôi
a) (x + 1/2)^2 = 1/16
=> (x + 1/2)^2 = (1/4)^2 hoặc (x + 1/2)^2 = (-1/4)^2
=> x + 1/2 = 1/4 hoặc x + 1/2 = -1/4
* x + 1/2 = 1/4
x = 1/4 - 1/2
x = -1/4
* x + 1/2 = -1/4
x = -1/4 - 1/2
x = -3/4
Vậy x = -1/4 hoặc x = -3/4
b) 2^x+2 - 2^x = 9^6
=> 2^x . 2^2 - 2^x = 9^6
=> 2^x . (2^2 - 1) = 9^6
=> 2^x . (4 - 1) = 9^6
=> 2^x . 3 = (3^2)^6
=> 2^x . 3 = 3^12
=> 2^x = 3^12 : 3
=> 2^x = 3^11
Vì 3^11 không chia hết cho 2
=> Không có giá trị nào của x thõa mãn đề bài
c) (3^x)^2 : 3^3 = 1/243
=> 3^2x = 1/243 . 3^3
=> 3^2x = 1/243 . 27
=> 3^2x = 1/9
=> 3^2x . 9 = 1
=> 3^2x . 3^2 = 1
=> 3^2x+2 = 1
=> 3^2x+2 = 3^0
=> 2x + 2 = 0
=> 2x = 0 - 2
=> 2x = -2
=> x = -2 : 2
=> x = -1
a\(\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2 \)
\(\Rightarrow x+\frac{1}{2}=\frac{1}{4}\)
\(\Leftrightarrow x=\frac{1}{4}-\frac{1}{2}\)
\(\Rightarrow x=\frac{-1}{4}\)